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Elasticity, Meaning (SPM Physics)

The property of a material that allows it to return to its original shape and size after the force deforming it is removed.

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Definition

The property of a material that allows it to return to its original shape and size after the force deforming it is removed.

Force and Motion II

What you need to know

Elasticity is the ability of a material to return to its original shape and size after a deforming force is removed. Within the elastic limit, the extension of a spring is directly proportional to the applied force, a relationship known as Hooke's law and written as F = kx.

Here, F is the applied force in newtons (N), x is the extension in metres (m), and k is the spring constant in newtons per metre (N m⁻¹), which describes how stiff the spring is. A larger spring constant means a greater force is needed to produce the same extension.

Beyond the elastic limit, the spring no longer returns to its original length even after the force is removed, and the relationship between F and x is no longer a straight line.

Stretching a spring stores energy in it as elastic potential energy, given by E = ½Fx or equivalently E = ½kx², measured in joules (J). This formula must be memorised, since it is not printed in the 4531 examination paper.

Worked example

A spring extends by 4 cm when a force of 8 N is applied to it, and this extension remains within the elastic limit of the spring. First convert the extension into metres: x = 4 cm = 0.04 m.

Using Hooke's law, F = kx, the spring constant is k = F ÷ x = 8 N ÷ 0.04 m = 200 N m⁻¹.

The elastic potential energy stored in the stretched spring is calculated using E = ½Fx = ½ × 8 N × 0.04 m = 0.16 J. The same answer is obtained using E = ½kx² = ½ × 200 N m⁻¹ × (0.04 m)² = 0.16 J.

Every step keeps consistent SI units throughout: force in newtons, extension in metres, spring constant in newtons per metre, and stored energy in joules, so no unit conversion errors are carried into the final answer. This consistent use of SI units throughout every step is exactly what examiners check for when awarding full marks in a calculation question of this type.

How it is examined

In Paper 1, objective questions often ask candidates to calculate extension, force or spring constant using F = kx, or to identify the region of a force–extension graph that represents the elastic limit. In Paper 2, structured questions typically ask candidates to state Hooke's law, define the elastic limit, and calculate elastic potential energy stored in a spring from given data or a graph.

In Paper 3, practical tasks commonly involve measuring the extension of a spring for a series of added weights, plotting a force–extension graph, and asking candidates to determine the spring constant from its gradient.

A common mistake is forgetting to convert extension from centimetres to metres before substituting into a formula, or using the total length of the stretched spring instead of the extension itself. Candidates should also remember that the elastic potential energy formula applies only within the elastic limit, and cannot be used once the spring has been permanently deformed.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Frequently asked questions

What is the difference between the elastic limit and the extension of a spring?
Extension is the increase in length of a spring measured in metres (m) when a force stretches it. The elastic limit is the maximum force, or corresponding extension, beyond which the spring no longer obeys Hooke's law and does not return to its original length once the force is removed.
Why does the spring constant k stay the same for small extensions?
Within the elastic limit, the extension of a spring is directly proportional to the applied force, so the ratio F ÷ x, which is the spring constant k, stays constant. This constant, measured in N m⁻¹, only changes if the spring is physically altered or stretched beyond its elastic limit.
Do F = kx and E = ½kx² give different information about a stretched spring?
Yes. F = kx relates the force applied to the resulting extension at any single instant. E = ½kx² instead gives the total elastic potential energy, in joules, stored in the spring due to that extension. Both formulas use the same spring constant k but describe different physical quantities.

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